Question: Let $h(x)=x^4-2x^3$. On which intervals is $h$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac32, \infty\right)$ only (Choice B) B $\left(-\infty,\dfrac32\right)$ only (Choice C) C $(-\infty,0)$ and $\left(\dfrac32, \infty\right)$ (Choice D) D $\left(0,\dfrac32\right)$ only (Choice E) E The entire domain of $h$
Answer: We can analyze the intervals where $h$ is increasing/decreasing by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $h$ is $h'(x)=2x^2(2x-3)$. $h'(x)=0$ for $x=0,\dfrac32$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=\dfrac32$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty,0)$ $(0,\frac{3}{2})$ $\frac{3}{2}$ $(\frac{3}{2},\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,0)$ $x=-1$ $h'(-1)=-10<0$ $h$ is decreasing $\searrow$ $\left(0,\dfrac{3}{2}\right)$ $x=\dfrac{1}{2}$ $h'\left(\dfrac12\right)=-1<0$ $h$ is decreasing $\searrow$ $\left(\dfrac{3}{2},\infty\right)$ $x=2$ $h'(2)=8>0$ $h$ is increasing $\nearrow$ In conclusion, $h$ is increasing over the interval $\left(\dfrac32, \infty\right)$ only.